Lesson video

Hello everyone and welcome to this trigonometry lesson.

I am Mr. Grattan, and in today's lesson, well, let's have a look at understanding the origins of the sine and cosine ratios from the sides of a right angled triangle.

Pause here to check that you are familiar with some of the really important keywords that we'll be using today.

Let's first look at triangles from the unit circle and see how we can use them along with ratio tables to find the lengths of sides of a triangle.

Izzy recalls that it is possible to find the value of X on this triangle since it is a triangle from the unit circle.

We know it's a triangle from the unit circle because its hypotenuse has a length of one unit, in this case one centimetre.

If we take an angle on any triangle from the unit circle such as this 35 degree angle, then the side opposite that angle will have a length of sine of that angle.

So in this case, sine of 35 degrees.

Furthermore, the side that's adjacent to both the angle, 35 degrees, and the right angle, so not the side that is the hypotenuse.

Well, that adjacent side has a length of the cosine of that angle, in this case cosine 35 degrees.

It's all good and well saying sine 35 degrees, but that represents a length.

So how can we find the actual length of sine 35 degrees? Well, we can use a table of trig values.

Let's have a look at the sine column and find the angle 35 degrees and look on over.

We can see that the length of sine 35 is 0.

574, so the length of the side opposite 35 degrees is 0.

574 centimetres.

We can then use this triangle from the unit circle.

This triangle with a hypotenuse of length one unit to generate similar triangles with hypotenuses either greater than or less than one unit.

So here's my triangle from before, a hypotenuse of one centimetre and a side opposite the angle 35 degrees with length 0.

574 centimetres.

Let's enlarge that triangle so its hypotenuse is 7.

3 centimetres instead.

What will the length of the opposite side be now? Well, I can generate this ratio table focusing on the hypotenuses.

I noticed that the scale factor multiplier from the triangle with the hypotenuse of one to its enlargement is a multiply by 7.

3.

And so the side opposite 35 degrees can also be multiplied by 7.

3 to find the length of the opposite side of the enlargement.

So 0.

574 times by the scale factor multiplier of 7.

3 gives an opposite side length of 4.

1902 centimetres.

But imagine we didn't have a triangle from the unit circle, one with a hypotenuse of one.

How can we find the lengths of the sides of, well, any right angled triangle, one with any lengthed hypotenuse? Jacob suggests we do exactly the same as before.

Focus on the ratio tables that show some enlargements.

Izzy is also right.

We can do this by first of all sketching a triangle with a hypotenuse of one, which is similar to our original given triangle because it'll have the same corresponding angles.

For example, we have this triangle whose hypotenuse is four centimetres and has an angle of 30 degrees.

Let's sketch a similar triangle whose hypotenuse is one centimetre but still has a 30 degree angle like so.

The side opposite that 30 degrees is sine of 30 and the side adjacent to the 30 degree angle is cosine of 30.

Notice how the corresponding side to A centimetres on our first triangle is the side on our sketch that is adjacent to 30 degrees, the one with the length cosine 30.

Using the table of trig values, we can spot the cosine of 30 degrees is 0.

866.

And so on our triangle from the unit circle, the adjacent side is 0.

866 centimetres.

We can now construct a ratio table of all the information we have from both triangles.

Starting with the sketch we have a hypotenuse of one and an adjacent side of 0.

866, and on our original triangle, which has a hypotenuse of four and an adjacent side of, well, we currently don't know, we can spot that the multiplier from the sketch to the original triangle is a multiplied by four.

And so the length of the side labelled A is 0.

866 multiplied by four, which gives you 3.

464 centimetres, and so on our original triangle, the side adjacent to the angle 30 degrees is 3.

464 centimetres.

The length four times larger than the 0.

866 on the triangle from the unit circle.

Okay, let's do a few checks with this triangle.

For the first check, pause here to answer which of these statements is correct for the side labelled P centimetres.

P is the side opposite that 20 degree angle.

Sticking with the same triangle, a similar triangle is sketched, one that has a hypotenuse of one.

Pause here to answer which side is corresponding to the side labelled P centimetres.

The side corresponding to P centimetres is the one labelled sine 20 degrees.

And now focusing on the sketch, pause here to use the table of trig values in order to find the length of the side labelled x centimetres on your sketch.

The length of the side labelled sine 20 degrees is sine 20, therefore 0.

342 centimetres.

And lastly pause here to use this ratio table to find the length of the side labelled P centimetres.

Okay, starting with A, A is the hypotenuse of our original triangle, which is 5.

6 centimetres, and B is the length of the side opposite the 20 degree angle in our sketch, that is the 0.

342 centimetre side.

And now for C, we look at the multiplier or scale factor between our sketch and our original triangle.

Let's focus on the hypotenuses to get from 1 to 5.

6.

That is a multiplier of times by 5.

6, therefore we can do 0.

342 times by 5.

6 to give us 1.

92 centimetres.

That is the length of the side labelled P centimetres.

Okay, onto the practise tasks, for question one, pause here to fill in the blanks on the triangle and in the ratio table by referring to the table of trig values by considering corresponding sides between the two triangles.

Find the length of the side labelled A centimetres.

And for question two, pause here to fill in all the blanks on all of the triangles and ratio tables and find the values of B and C by considering corresponding sides.

And for question three you'll have to sketch a similar triangle for yourself.

Each sketch has to have a hypotenuse of one unit.

Use these sketches to find the length of the missing side of each of these four triangles.

So onto the answers, pause here to check all of the values that you filled in and compare them with the ones on screen.

And well done if you found that the side labelled A centimetres has a length of 11.

592 centimetres.

For question two, pause it and check that all of your values match with the ones on screen and are very well done.

If you found that the length B centimetres is 6.

1425 centimetres and C centimetres is 6.

156 centimetres.

And for question three, the length of the side A centimetres is 6.

5 centimetres, B centimetres is 21.

744 centimetres.

C centimetres is the length 6.

7408 centimetres and side D centimetres has a length of 0.

4979 centimetres.

Okay, let's see if we can spot a relationship between the side opposite an angle and the hypotenuse of a triangle.

Izzy thinks that she spotted a pattern.

If we have two similar triangles where one of them is from the unit circle and has a hypotenuse of one unit, then the scale factor between them is always the hypotenuse of that other triangle.

This is an outstanding observation.

And Jacob is correct also.

All the sides of the triangle with a hypotenuse of one centimetre are multiplied by the same scale factor, the hypotenuse of that other triangle in order to get the lengths of the corresponding sides of that other triangle.

So let's have a focus on the triangle with a hypotenuse of one centimetre.

The length of the side opposite the 35 degree angle is sine 35, but we know that every side length on that triangle is multiplied by 7.

3 to get the corresponding side on the other triangle.

Therefore, the length of the side currently labelled Y is sine 35 degrees times by 7.

3.

The pattern Izzy noticed works for all right angle triangles that is similar to one with hypotenuse of one unit whose side opposite and angle is sine theta.

This similar triangle can have hypotenuse of any length.

Let's label it H centimetres, and any angle, feet or degrees, will work as long as it is sensible to go into a right angle triangle.

If all of this is true, then this side opposite the angle theta will be the hypotenuse times by sine theta, and here's how.

If this is sine theta, then the multiplier or scale factor is going to be times by H.

And so the corresponding side to sine theta, which is the side opposite the angle theta, is then multiplied by the hypotenuse.

This is of great benefit.

Rather than having to sketch a similar triangle for every single question, we can instead simply look at the relationship between two sides of one triangle.

There is a multiplicative relationship between the hypotenuse and the opposite sides on all right angled triangles.

I take the hypotenuse and multiply it by sine theta.

This always gives the length of the side opposite to theta.

H times sine theta or the hypotenuse times sine theta.

For this check, pause here to use this relationship in order to find the values that should go into these two gaps.

I start with the hypotenuse of 46 centimetres and multiply it by sine 75 to get the length of the opposite side.

Jacob asks, we've just seen that we can find the length of a side opposite an angle if we know the length of the hypotenuse and the size of the angle itself.

But what if instead we know the length of the opposite side but not the hypotenuse? Can we still find the length of the hypotenuse using a similar method? And the answer is we can, put simply, we can substitute the information that we do know into our relationship and then rearrange and solve the equation we get after our substitution.

For example, this is our relationship.

We know that the side opposite 35 degrees is 10 centimetres and we know that the angle is well, 35 degrees.

We can use our table of trig values to spot that sine 35 equals 0.

574.

Now we have an equation that we can rearrange by dividing through both sides by 0.

574 to get H equals 10 over 0.

574.

Okay, for this check, pause here to use the relationship to find the values that should go into these two gaps.

The side opposite angle 75 degrees is 46.

So 46 equals the unknown hypotenuse length times by the sine of 75.

Pause here to use your table of trig values to answer what goes into these two gaps when trying to rearrange and solve this equation.

Using our table of trig values, we can see that sine 75 has a value of 0.

966.

Okay, here's the next set of practise questions.

For all parts of question one, find the length of the side that is opposite the labelled angle using this relationship, and for parts A and B, fill in every gap in your method to help you find those lengths.

Pause now for question one.

And for question two it is the hypotenuse that is unknown instead of the opposite side.

So pause here to substitute, rearrange, and solve the equations created from our relationship between the opposite side and the hypotenuse for each of these four triangles.

And here are the answers to question one.

Pause now to compare your method steps and lengths of opposite sides to the ones that you can see on screen.

And for question two, pause here to compare your method steps and lengths of hypotenuses to those on screen.

So we've looked at the relationship between the opposite side and the hypotenuse.

Let's see if there is a similar relationship between the adjacent side and hypotenuse instead.

And, in fact, there is.

Like with before, I have a triangle with a hypotenuse of one centimetre and a similar triangle with any hypotenuse H.

But this time the sides adjacent to the angle theta are labelled rather than the opposite ones from before.

On any right angled triangle, the side adjacent to the angle theta will be H multiplied by the cosine of theta.

In other words, the hypotenuse multiplied by cosine theta, and here's how.

The scale factor between the two triangles is still H, the hypotenuse of that other triangle.

If the side adjacent to theta on my triangle from the unit circle is cos theta, then that corresponding side, the side adjacent to the angle theta on the other triangle has a length of cos theta times by H.

Once again, rather than having to sketch a similar triangle for every question, we can look at the relationship between these two sides of one of our triangles.

There is a multiplicative relationship between the hypotenuse and the adjacent side on all right angled triangles.

I take the hypotenuse and multiply it by cosine theta.

This always gives me the length of the side adjacent to theta.

H times by the cosine of theta.

The relationship between these two sides can be shown by the adjacent side is equal to the hypotenuse times by cosine of theta.

Okay, let's have a look at two examples.

For this first triangle, we know the hypotenuse of 22 centimetres and an angle of 70 degrees.

We know the side adjacent to the 70 degrees will be H times by cosine of theta, which is 22 centimetres for the hypotenuse times by cosine of 70 degrees.

Using our table of trig values, we can say that we've got 22 times by 0.

342.

And so the side adjacent to the 70 degree angle and the right angle is 7.

524 centimetres long.

Furthermore, for this second triangle, we know that there's an angle of 70 degrees and a side adjacent to it with a length of 22 centimetres.

Notice how the side of length 22 centimetres is a different part of the two triangles and therefore these two triangles are not congruent.

Once again, I start with my relationship, the adjacent equals H times by cosine of theta, but this time I substitute the 22 into the adjacent variable whilst the 70 degrees remains as the substitution for the theta variable.

We still know that cosine of 70 is 0.

342, but now we've got the equation 22 equals H times by 0.

342.

From here we can rearrange the equation to solve it by dividing through by 0.

342, giving us H equals 22 divided by 0.

342, which is 64.

33 when rounded.

So in conclusion, if we are trying to find the length of the hypotenuse, then a division is required, whilst when trying to find the adjacent to an angle, division is not required.

Okay, for this check, pause now to match the statement to the angles marked X and Y.

The side adjacent to the 75 degree angle is Y, whilst the hypotenuse of that triangle is X.

Neither X nor Y was a side opposite to the 75 degree angle.

And for this next check pause here to use this relationship to find the values that should go into these two gaps.

We take the hypotenuse of six centimetres and multiply it by the cosine of the angle 75 degrees to give the length of the side that is adjacent to the angle 75 degrees.

And for this final check pause here to use this relationship to find the values that go into these two gaps instead.

We do not know the length of the hypotenuse, but we do know that the length of the side adjacent to 75 degrees is six centimetres long.

And now we want to rearrange and solve this equation in order to find the length of H, the hypotenuse.

Pause here to answer what would go into these two gaps in our method in order to solve this equation.

And you can use your table of trig values to help you out.

Using our table of trig values, cos(75) is 0.

259.

Okay, for these practise questions, for all parts of question one, find the length of the side that is adjacent to the labelled angle using this relationship.

And for parts A and B, fill in every gap in the method to help you find those lengths.

Pause now for question one.

And for question two, it is the hypotenuse that is unknown instead of the adjacent side.

Pause here to substitute, rearrange, and solve the equations created from the relationship between the adjacent side and the hypotenuse for each of these four triangles.

And here are the answers to question one.

Pause now to compare your method and lengths of the adjacent sides to those that are on screen.

And for question two, pause here to compare your method steps and lengths of hypotenuses to those that you can see on screen.

We've been using the table of trig values to answer all of our questions so far, but are there limitations to only using a table of trig values? Let's have a look.

So far, Jacob seems to think that the table of trig values is pretty useful for this triangle.

He knows that K is the length of the adjacent and the hypotenuse has a length of 104 centimetres.

He can use all of this information to calculate the length of the side adjacent to the 55 degree angle using this table of trig values because he needs it to find the value of cosine 55.

So using that table of trig values, we've got cosine 55 is 0.

574.

We can then substitute 0.

574 in place of cosine 55 to give us 104 times by 0.

574 for a total length of K of 59.

70.

But now Jacob is stuck for this very similar question.

There is no way of him finding what cosine 57 degrees is since it's not mentioned anywhere on the table.

That is where your calculator can come in.

It can easily find the sine and the cosine of any angle using these two trig function buttons.

But first, in order for your calculator to get some accurate results, we need to make sure your calculator is accepting angles that are measured in degrees rather than other angle measurements such as radians.

You don't need to know about those.

To do this, let's have a look at the settings button and click right to choose calculator settings.

From there, scroll down to the angle unit option, click right, and make sure you have chosen the degrees option, not the radians or gradians option.

When you've correctly selected the degrees option, click okay to go back to your normal calculator menu.

Okay, now that you know your calculator will work with angles measured in degrees, let's use the calculator to find the length of the side adjacent to the right angle and that 57 degree angle.

First we start with the length of the hypotenuse of 123 centimetres.

Then we multiply it by the cosine.

Oh, and note that the open bracket does appear by itself.

Of the angle adjacent to the side we want to find the length of, in this case, the 57 degrees.

Because our calculator is already set to degrees mode, there's no need to find and put in the degrees symbol.

Your calculator automatically knows that the angle is in degrees.

Make sure that you also type in the close bracket button yourself.

And then when you have 123 multiplied by cosine 57, close bracket, press execute, which will give you your answer, the length of the side adjacent to the 57 degree angle.

For this check, pause here to use a calculator in order to find the lengths of the two sides of this triangle labelled A and B, rounding your answer to three decimal places.

The length of the side adjacent to the 32 degree angle is 0.

848, whilst the length of the side opposite the 32 degree angle is 0.

530 centimetres.

And similar again, but this time you have to figure out on your own which sides are the sine and the cosine of the angle, in this case 53 degrees.

Pause now to use your calculator to find the lengths of C and D.

Remember the opposite is always sine and the adjacent is always cosine.

And so we have cosine of 53 degrees equals 0.

602 and the sine of 53 degrees is 0.

799.

And similar checks again.

But this time the hypotenuse of each triangle is not one unit.

Pause now to use the relationships that we've looked at today and your calculator to find the lengths of the adjacent and opposite sides A and B.

The answers are 7.

757 centimetres and 13.

994 centimetres.

And finally, use a calculator in order to find the length of the hypotenuse of this triangle.

You will have to solve an equation in order to answer this question.

Oh, and which relationship can you use for this question? The sine ratio involving the opposite side or the cosine ratio involving the adjacent side? Pause here to choose and to answer this last check question.

You would've had to type into your calculator 49 divided by sine of 64 to get 54.

5 centimetres.

Okay, onto the final set of practise questions.

For question one, pause here to use your calculator in order to group these cards by matching their values.

Some groups will have two cards and some will have three.

And finally, question two, use a calculator to find the lengths of all missing sides on each of these triangles.

Every triangle is only slightly different from the last.

So see if you can be careful and efficient with how you use your calculator.

Pause now to do all six of these questions.

And onto the answers.

Here are the four groups for question one.

Do you spot anything about the angles in the sine and cosine functions in the two groups of three? Pause now to have a look at your answers and discuss this extra question.

And finally pause here to check your answers for the lengths of all of the sides and see if they're correct when compared to the ones on screen.

And thank you all for all of your hard work and effort in this really packed lesson as we've taken a journey through the sine and cosine ratios by looking at triangles from a unit circle whose hypotenuse is one unit and whose side adjacent and opposite to an angle theta are cosine theta and sine theta respectively.

We've also used a table of trig values to consider the values of sine and cosine theta, but only for certain angles.

And finally, we've spotted and used relationships between different pairs of sides of the triangle, which are the opposite equals the hypotenuse times by sine theta, and the adjacent equals the hypotenuse times by the cosine of theta where sine and cosine of any angle can be found using a calculator.

Once again, thank you all for joining me for this lesson, and until our next math lesson together, take care and have an amazing rest of your day.